Shear EC2 UKNA SPREADSHEETS Structural Design Carlo Sigmund Shear IN ACCORDANCE WITH European Standards CEN/TC 250 Structural Eurocodes (EN 1992-1-1) UK National Annex ... Contents

EUROCODESS P R E A D S H E E T SS t r u c t u r a l D e s i g n

C a r l o S i g m u n d

ShearI N A C C O R D A N C E W I T H

E u r o p e a n S t a n d a r d s C E N / T C 2 5 0S t r u c t u r a l E u r o c o d e s ( E N 1 9 9 2 - 1 - 1 )

UK National Annex

Es

E b o o kFirst Edition

Copyright© 2014 http://eurocodespreadsheets.jimdo.com

All rights reserved. Carlo Sigmund

------------------------------------------------

First Edition: April 2014

Author/Publisher: Sigmund, Carlo <1971->

ISBN n.:

ID contenuto:

Ebook: Shear EC2

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The sponsoring editor for this document and the production supervisor was Carlo Sigmund.

Although care has been taken to ensure, to the best of our knowledge, that all data and information contained herein are accurate to the extent that they relate to either matters of fact or accepted practice or matters of opinion at the time of publication, The EUROCODES Spreadsheets Structural Design, the author and the reviewers assume no responsibility for any errors in or misinterpretations of such data and/or information or any loss or damage arising from or related to their use.

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Contents

Shear_EC2_UKNA................................................................................................3

1.1 General: ShearReinforcementBeamSlab.xls ........................................................................................... 3

1.2 Layout ...................................................................................................................................................... 4

1.3 Output - Word document (calculation sheet)............................................................................................ 7

1.4 Shear_EC2 (Beams and slabs) derived formulae.................................................................................... 8

1.5 Verification tests....................................................................................................................................... 10

1.6 References............................................................................................................................................... 14

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Topic: User’s Manual/Verification tests - ShearReinforcementBeamSlab.xls page 3

Section 1 Shear_EC2_UKNA

1.1 General: ShearReinforcementBeamSlab.xls

his sheet allows for the design of a section of solid slab or a rectangular beam section. The spreadsheet checks beams or slabs for shear and calculates any

shear reinforcement required in accordance with EN 1992-1-1:2004 Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings and BS NA EN 1992-1-1 (2004) (English): UK National Annex to Eurocode 2. Design of concrete structures. General rules and rules for buildings.

The user is required to input desired characteristic strength of the materials, amount of concrete cover to shear reinforcement ( - EN 1992-1-1 Exp. 4.1) and the longitudinal reinforcement ( ) which extends > beyond the section considered. All applied load and stress should be ultimate and positive.

With regard the concrete, the minimum strength used is C12/15 (Table 3.1). The spreadsheet allows input for grades of concrete up to C50/60. Input is self explanatory but, in order to facilitate use of this spreadsheet, some degree of automation has been introduced.

The design of members with shear reinforcement is based on the truss model given in Figure 6.5 (EN 1992-1-1). The angle between the concrete compression strut and the beam axis perpendicular to the shear force is chosen within the range 1,00÷2,50 (Cl. 6.2.3(2) Exp. (6.7N)).

Applied shear force ( ) is compared with three values for the resistance ( ). represents the shear capacity of concrete alone; is the shear

resistance determined by the capacity of the notional concrete struts; and is the capacity of a section with shear reinforcement.

Where more than one vertical links are required per section, the user may choose to use inner links alternative or bent-up bars in order to increase section strength. The spreadsheet takes automatic measures to ensure maximum spacing (longitudinal and transversal) of shear reinforcement is met (Cl. 9.2.2 for Beams and Cl. 9.3.2 for Solid slabs).

Combo-boxes to the right under “DETAILING” Excel® Form define minimum bar sizes to be used. Combo-box to the right under “PreCalculus” Excel® Form define concrete characteristic compressive strength at 28 days.

T

cnomAsl lbd d+

VEd VRdVRd C VRd max

VRd s

EUROCODES SPREADSHEETS STRUCTURAL DESIGN SECTION 1 SHEAR_EC2_UKNA

page 4 Topic: User’s Manual/Verification tests - ShearReinforcementBeamSlab.xls

No check is performed in order to control cracking of the concrete in shear. Therefore the maximum shear force allowed is checked (Exp. 6.5, Cl. 6.2.2(6)).

The value of shear force, , used can, provided there is diagonal compression and continuity of tension reinforcement ( ) for at least (design anchorage length), be evaluated at “d” from the face of support (see Figure 6.3).

This method presented in EC2 is known as “The Variable Strut Inclination Method”. The use of this method allows the designer to seek out economies in the amount of shear reinforcement provided, but recognising that any economy achieved may be at the expense of having to provide additional curtailment and anchorage length to the tension steel over and above the normally required for resistance to bending. Calculations with axial forces in the cross-section due to loading or prestressing are not carried out in this spreadsheet: with

.

1.2 Layout

The basic layout for Shear_EC2 sheet is shown on the following screen dump:

VEdAsl lbd

NEd 0=cp 0=

Figure 1.1 Sheet “Splash” from Excel® 2003.


EUROCODES SPREADSHEETS STRUCTURAL DESIGNSECTION 1 SHEAR_EC2_UKNA

DETAILING Form:

Figure 1.2 Input (white blank test boxes) and calculus (blue text boxes). Default value for cc = 0,8 (red).

Figure 1.3 Report.



PreCalculus Form:

View Notes...

How to remove links to Excel file “ShearReinforcementBeamSlab.xls”:

Figure 1.4 Coloured in blue results in output. White text boxes only to change the design values.



1.3 Output - Word document (calculation sheet)

Example calculation sheet (output):

Figure 1.5 Sample print with compression steel (Word® Office® must be installed on your system).



1.4 Shear_EC2 (Beams and slabs) derived formulae

The following notation is used in the equations for the shear design:

• the shear force due to the actions at the ultimate limit state occurring at cross-section being considered (see Figure 6.3)

• is the design value of the maximum shear force which can be sustained by the member, limited by crushing of the compression struts

•

• is the angle between the concrete compression strut and the beam axis perpendicular to the shear force

• is the angle between shear reinforcement and the beam axis perpendicular to the shear force

• effective depth of the cross-section

• width of the web on T, I or L beams

• characteristic yield strength of shear reinforcement

• design yield of shear reinforcement

• is the cross-sectional area of the shear reinforcement

• is the longitudinal spacing of the shear reinforcement bars

• characteristic compressive cylinder strength of concrete at 28 days

• design value of concrete compressive strength

• is the coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied

• is a coefficient taking account of the state of the stress in the compression strut

• is a strength reduction factor for concrete cracked in shear.

Shear reinforcement is not normally required provided the design ultimate shear force does not exceed :

(Eq. 1‐1)

but not less than:

(Eq. 1‐2)

where:

and . Where is the area of tensile reinforcement, which extends > beyond the section considered (see EN 1992-1-1, Figure 6.3).

VEd

VRd max VRd max tan;cot =

VRd s VRd s tan;cot =

d

bw

fywk

fywd fywk 1 15 0 87fywk= =

Asw

s

fck

fcd ccfck 1 50=

cc

cw

1 0 6 1 fck 250– 1 0 5 cos– = =

VEd VRd C

VRd C CRd C k 100lfck 1 3/ bwd=

vminbwd 0 035k3 2/ fck1 2/ bwd=

k 1 200 d mm + 2 0= l Asl bwd 0 02= Asllbd d+



When, on the basis of the design shear calculation, no shear reinforcement is required, minimum shear reinforcement should nevertheless be provided according to 9.2.2.

Where exceeds shear reinforcement is required.

The beam section should be considered inadequate whether at least one of the following two conditions is not met:

(Eq. 1‐3)

. (Eq. 1‐4)

In such case the section should be resized or the materials strength increased.

If Eq. 1-3 and 1-4 are met:

a. if the design value of the shear force which can be sustained by the yielding shear is given by

b. if calculate angle , of strut for full shear force at end of beam

c. if the design value of shear force is given by

d. if the design value of the shear force is given by: .

The shear force applied to the section must be limited so that excessive compressive stresses do not occur in the diagonal compressive struts, leading to compressive failure of the concrete. Thus the maximum design shear force

is limited by the ultimate crushing strength of the diagonal concrete member in the analogous truss and its vertical component. According to EN 1992-1-1:

• for members with vertical shear reinforcement:

(Eq. 1‐5)

• for members with inclined shear reinforcement:

(Eq. 1‐6)

where is the lever arm due to bending with shear.

In the case of “section adequate” with , all shear will be resisted by ( ) the provisions of shear reinforcement bars with no direct contribution from the shear capacity of the concrete itself. According to EN 1992-1-1:

• for members with vertical shear reinforcement:

VEd VRd C

VEd 0 5bwdfcd

VEd VRd max VRd max = cot 1 00 =

VEd VRd max VRd max = cot 2 50 =

VRd s VRd s = cot 2 50 =

VEd VRd max VRd max = cot 2 50 = 0 5arc 2VEd cw1fcd bwz sin=

21 80 45 VRd s VRd s cot =

21 80VRd s VRd s = cot 2 50 =

VRd max

VRd max cwbwz1fcd

cot tan+ -----------------------------------=

VRd max cwbwz1fcdcot tan+

1 cot2+ ------------------------------------=

z 0 9d=

VEd VRd CVRd s



(Eq. 1‐7)

• for members with inclined shear reinforcement:

(Eq. 1‐8)

where is the lever arm due to bending with shear. For bent-up bars according to the truss model in Figure 6.5 the longitudinal spacing between shear assemblies is given by:

(Eq. 1‐9)

where “n” is the number of shear assemblies along the beam axis. The area of each shear assemblies is . In the case of only one shear assembly (n = 1) the longitudinal spacing to be considered in the shear calculations is given by:

(Eq. 1‐10)

where .

1.5 Verification tests

SHEARREINFORCEMENTBEAMSLAB.XLS. 5.58 MB. Created: 26 December 2013. Last/Rel.-date: 04 November 2015. Sheets:

— Splash

— CREDITS.

EXAMPLE 1-A‐ Pre‐dimensioning and checking shear reinforcement (UK National Annex to EC2) ‐ test1

Given: The beam cross‐section in Figure below is given. The characteristic material strengths are C30/37 for the concrete and for the steel. Check if the shear reinforcement in the form of vertical link ( %) and bent‐up bars can support, in shear, the given ultimate load .

Solution: Take an effective depth . The breadth of section . The nominal cover to links . Level arm .

Design shear stress occurring at cross‐section being considered:

.

Design materials strengths:

concrete

VRd sAsws

---------zfywd cot=

VRd sAsws

---------zfywd cot cot+ sin=

z 0 9d=

szn--- cot cot+ =

Asw

s z cot cot+ =

45 90

fywk 500 MPa=3 50=

VEd 340 kN=

d 557 mm 550 mm= b bw 350 mm= =cnom 25 mm= z 0 9d 495 mm= =

vEd VEd bwd 340 103 350 550 1 77 MPa= = =

fcd ccfck 1 50 0 85 30 1 50 17 00 MPa= = =



,

, , (chosen).

.

Steel (shear reinforcement):

.

Area of tensile reinforcement which extends > ( ) beyond the section considered:

(considered 3H16 at the face of beam support).

SHEAR RESISTANCE (WITHOUT SHEAR REINFORCEMENT).

Design value for the shear stress resistance:

(satisfactory).

(satisfactory).

.

Figure 1.6 Beam cross-section.

fcd 0 6 1 30 250– 17 0 528 17 8 976 MPa= = =

stirrups 90= bentupbars 45= * MIN stirrups bentupbars; 45= =

1 1 0 5 *cos– 0 528 1 0 5 45 cos– 0 341= = =

fywd fywk 1 15 500 1 15 435 MPa= ==

lbd d+

Asl 600 mm2=

lAslbwd--------- 600

350 550 ---------------------------- 0 0031 0 02= = =

k 1 200d

---------+ 1 200550---------+ 1 603 2 0= = =

CRd C k 100lfck 1 3/ 0 12 1 603 100 0 0031 30 1 3/ 0 40 MPa= =

vmin 0 035k3 2/ fck1 2/ 0 035 1 603 3 2/ 30 1 2/ 0 39 MPa= = =

vRd C max 0 40 0 39; 0 40 MPa= =



Shear reinforcement required: .

(satisfactory)

with:

(beam adequate).

DETAILING. Shear reinforcement for beam (stirrups and bent‐up bars) with (see Cl. 9.2.2(4)). Design stress at cross‐section being considered

(with @stirrups and @bent‐up bars). Chosen with (Cl. 6.2.3(2) ‐ Note ‐ Exp. (6.7N)).

Stirrups: N = 1 enclosing links with diameter H10, and no inner links. Longitudinal

spacing (vertical links): (chosen). angle between vertical links and

the beam axis). ,

.

Bent‐up bars: 2H16 ( ) bent‐up at an angle from the bottom of the beam (for 2 times) with a longitudinal spacing given by:

.

Maximum longitudinal spacing (vertical links ):

.

Compression longitudinal reinforcement which is included in the resistance calculation (bending): H20 with .

(satisfactory).

Transverse vertical spacing:

where is the number of inner links.

Minimum shear reinforcement (for ):

(satisfactory).

vEd 1 77 vRd

·C

0 40 MPa= =

vEd 0 5fcd 0 5 0 6 1 30250---------–

17 4 49 MPa= =

VRd maxcwbwz1fcd1 tan+

------------------------------ 1 350 495 0 341 17 1 1+

---------------------------------------------------------- 502165 N= = =

vRd max vRd max = 1cot 502165350 550

---------------------------- 4 10 MPa vEd 1 77 MPa= = = =

3 0 5=vEd VEd bwd 1 77 MPa= =

0 5 1 77 0 88 MPa= 0 88 MPacot 1 00= 1 00 2 50

s 190 mm= 90=

Asw 2 78 157 mm2= =

Asw s 157 190 0 827 mm2 mm 827 mm2 m= = =

Asw 402 mm2= 45=

szn--- cot cot+ 495

2--------- 1 1+ 495 mm= = =

90=

sl max 0 75d 1 cot+ 0 75 550 1 0+ 413 mm= = =

sl max 15 20 mm 300 mm= =

min 413 300; 300 mm s 190 mm= =

stbw 2cnom– Enclink–NinnerLink 1+

------------------------------------------------- 350 2 25 – 10–0 1+

----------------------------------------- 290 mm= = =

NinnerLink

sin 1=

Asws

---------

minw min bw sin

0 08 fckfywk

---------------------- bw 1 0 08 30500

---------------------- 350 1 0 306 mm2

mm----------- 306

mm2

m-----------= = = = =

Asws

--------- 0 827 mm2 mmAsws

---------

min=



SHEAR CAPACITY. Capacity of the concrete section with vertical and inclined shear reinforcement to act as a strut:

(vertical links)

(bent‐up bars).

Value used for shear calculations:

.

SHEAR REINFORCEMENT (VERTICAL LINKS):

, .

.

(satisfactory).

SHEAR REINFORCEMENT (BENT‐UP BARS):

, , .

.

(satisfactory).

NOTE. Provide H10 mm diameter enclosing links, (and no inner link) at 190 mm centres. Provide 2H16 bent‐up at for times with a longitudinal spacing

.

MAXIMUM EFFECTIVE CROSS‐SECTIONAL AREA OF THE SHEAR REINFORCEMENT (COT = 1,00):

Stirrups:

VRd maxcwbwz1fcd

cot tan+ ----------------------------------- 1 350 495 0 341 17

1 1+ -------------------------------------------------------------- 502165 N= = =

VRd maxcwbwz1fcd1 cot2+

------------------------------ cot cot+ 1 350 495 0 341 17 1 12+

-------------------------------------------------------------- 1 1+ 1004330 N= = =

VRd max min 502165 1004330; 502165 N= =

Asws

--------- 0 827 mm2 mm=

cot 1 00= cot tan+ 2 00=

VRd sAsws

---------zfywd cot 0 827 495 435 1 00 178074 N= = =

min VRd s VRd max; bwd min 178074 502165; 350 550

------------------------------------------------------- 0 92 Mpa vEdLink 0 88 MPa= = =

Asws

--------- 2 201 495

--------------------- 0 812 mm2 mm= =

cot cot+ 2 00= sin 45 sin 0 71= = 1 cot2+ 2 00=

VRd sAsws

---------zfywd cot cot+ sin 0 812 495 435 2 00 0 71 248278 N= = =

min VRd s VRd max; bwd min 248278 502165; 350 550

------------------------------------------------------- 1 29 Mpa vEdLink 0 88 MPa= = =

45= n 2=s 495 mm

Asw max

s------------------ 0 5cw1

fcdfywd---------bw 0 5 1 0 341 17

435 ------------- 350 2 332 mm2 mm= =



Bent‐up bars:

.

1.6 References

BRITISH STANDARDS INSTITUTION. The structural use of concrete – Part 1: Code of practice for design and construction, BS 8110–1:1997

EN 1992-1-1:2004. Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings. Brussels: CEN/TC 250 - Structural Eurocodes, December 2004 (DAV)

EN 1992-1-1:2004/AC:2010. Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings. Brussels: CEN/TC 250 - Structural Eurocodes, December 2010Reinforced Concrete Design to Eurocode, Bill Mosley, John Bungey and Ray Hulse, Palgrave Macmillan. 7th Edition

BS NA EN 1992-1-1 (2004) (English): UK National Annex to Eurocode 2. Design of concrete structures. General rules and rules for buildings

Reinforced Concrete Design to Eurocode, Bill Mosley, John Bungey and Ray Hulse, Palgrave Macmillan. 7th Edition.

Asw max

s------------------ 0 5cw1

fcdfywd sin----------------------bw 0 5 1 0 341 17

435 0 71 ------------------------------- 350 3 285 mm2 mm= =

Documents

Shear EC2 UKNA SPREADSHEETS Structural Design Carlo Sigmund Shear IN ACCORDANCE WITH European Standards CEN/TC 250 Structural Eurocodes (EN 1992-1-1) UK National Annex ... Contents